# Inverse problem with normal distributions and max

Consider $$n$$ independent normally distributed random variables

$$X_i\sim N(\mu_i,\sigma_i^2)$$

and denote $$Y = \max\limits_{1\leq i\leq n}\{X_i\}$$. We can define the probabilities, for each $$1\leq i\leq n$$,

$$\tilde{P}_i = \mathbb{P}(X_i=Y)$$

where $$\tilde{P}_i$$ is a function of $$\vec{\mu} = (\mu_1,\ldots,\mu_n)$$ and $$\vec{\sigma}=(\sigma_1,\ldots,\sigma_n).$$ I'm interested in the inverse problem: given some observed probabilities $$P_1,\ldots,P_n$$ (summing to unity),

$$\min\limits_{(\vec{\mu},\vec{\sigma})\in\mathbb{R}^{n\times n}}\sum_{i=1}^n|P_i - \tilde{P}_i|^2.$$

• I believe you can solve this exactly (that is, the minimum is $0$) by assuming $\sigma_i^2=1$ for all $i.$ An induction on $n$ ought to work, so--to understand the issue--start with the case $n=2.$ Indeed, you don't need many details: a topological argument will do, relying simply on the continuity of the function that maps the means $(\mu_i)$ to the probabilities $(\tilde P_i).$ – whuber Nov 28 '20 at 16:23
• I was thinking actually that it's not well-posed even with fixed unit variances. My reasoning is the following: for any solution $\mu^\star$ if we translate by a vector $(a,\ldots,a)$ then the probabilities $\tilde{P}_i$ would all remain the same. In fact, it seems to me that $\tilde{P}_i$ is translation invariant in $\vec{\mu}$ even without unit variance. – Dennis Nov 28 '20 at 16:52
• For the $n = 2$ case we can evaluate $\tilde{P}_i$, but this seems non-trivial for the case where $n > 2$, is that right? If you could upvote the question it would be appreciated. – Dennis Nov 28 '20 at 19:20